If one is undefined, shouldn't the other be undefined, too? They are inverse functions. For instance, since we have that

$$\lim_{x \to \frac{\pi}{2}} \tan x = \text{undefined}$$

so too should $$\lim_{x \to \infty} \arctan x = \text{undefined}$$

since there is no value for which $\tan x$ takes infinity.

  • 1
    $\begingroup$ Where do you see it defined? $\endgroup$
    – badjohn
    May 19 '20 at 16:08
  • $\begingroup$ how do you mean? $\endgroup$
    – 666User666
    May 19 '20 at 16:08
  • $\begingroup$ You say "limit of $\arctan(x) $at x$=\infty$ defined" where do you see that? $\endgroup$
    – badjohn
    May 19 '20 at 16:11
  • 2
    $\begingroup$ The confusion may be due to how $\infty$ is used in limits. It might look as if we are treating $\infty$ as a number but limits involving $\infty$ have their own definitions which don't actually involve $\infty$. It is just a suggestive shorthand. $\endgroup$
    – badjohn
    May 19 '20 at 16:13
  • $\begingroup$ The points at which $\tan$ is undefined correspond to vertical asymptotes. When you flip the graph to get the inverse, these vertical asymptotes become horizontal. $\endgroup$
    – amd
    May 19 '20 at 22:26

The limit of tangent at $\frac\pi2$ is undefined, but the limits approaching from below and above are defined: $$\lim_{x\rightarrow\frac\pi2,\;x<\frac\pi2}\tan(x)=+\infty, \\ \lim_{x\rightarrow\frac\pi2,\;x>\frac\pi2}\tan(x)=-\infty.$$ In particular, if we restrict the function $\tan$ to the open interval $(-\frac\pi2,\frac\pi2)$, then $\tan$ has limits at both ends.

The important thing to know is that $\arctan$ is defined as the inverse of this restricted function, not the general function on $\Bbb R$ (which is not even bijective). Therefore, nothing prevents $\arctan$ from having limits at $-\infty$ and $+\infty$, and in fact $$\lim_{x\rightarrow+\infty}\arctan(x)=\frac\pi2, \\ \lim_{x\rightarrow-\infty}\arctan(x)=-\frac\pi2.$$


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